# If n=110 and widehat{p} (p-hat) =0.66, construct a 95% confidence interval.Give your answers to two decimals? <p< ?

If $n=110$ and $\stackrel{^}{p}$ (p-hat) $=0.66$, construct a $95\mathrm{%}$ confidence interval. Give your answers to two decimals $?

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Step 1 The confidence interval for a sample proportion at $95\mathrm{%}$ can be calculated using the formula, $\stackrel{^}{p}-z\cdot \sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{N}}, where is the sample proportion, N is the sample size and z* is the approprite value from the standard normal distributions for the desired confidence intervals. The z* value for $95\mathrm{%}$ confidence interval is equal to 1.96 Step 2 The sample proportion is equal to 0.66 and the sample size is equal to 110 and the z* value is equal to 1.96 Substitute the values of $\stackrel{^}{p}$, N and z* in the formula, $\stackrel{^}{p}-z\cdot \sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{N}} to calculate the confidence interval of the sample proportion at $95\mathrm{%}$ interval $\stackrel{^}{p}-z\cdot \sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{N}}
$0.66-\left(1.96\sqrt{\frac{0.66\left(1-0.66\right)}{110}}\right)
$0.66-\left(1.96\sqrt{0.00204}\right)
$0.66-\left(1.96\left(0.0451\right)
$0.66-0.0884
$0.5716 The confidence interval for the sample proportion at $95\mathrm{%}$ is equal to (0.5716, 0.7484)